Question

If three vectors 2i – j - k, i + 2j – 3k and 3i + $\lambda j$ + 5k are coplanar, then the value of $\lambda $ is?
(a) -4
(b) -2
(c) -1
(d) -8

Answer 1

Hint: To solve this problem, we use the formula for scalar product on these vectors. The formula for scalar product of 3 vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$, is given by $\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})$. Here, $(\overrightarrow{b}\times \overrightarrow{c})$ denotes the vector product of two vectors. Since, the scalar product is 0 for coplanar vectors, we use this information to solve the problem.

Complete step-by-step answer:
Now, we need to find the condition for which three vectors 2i – j - k, i + 2j – 3k and 3i + $\lambda j$ + 5k are coplanar, thus, we will apply the formula for the scalar product of 3 vectors. Thus, we have,
= $\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})$
In this case, $\overrightarrow{a}$ = 2i – j – k, $\overrightarrow{b}$ = i + 2j – 3k, $\overrightarrow{c}$ = 3i + $\lambda j$ + 5k, thus, we have,
= \[\left( 2ijk \right).\left( \left( i+2j3k \right)\times \left( 3i+~\lambda j+5k \right) \right)\] -- (A)
We will use the matrix notation for calculating the vector product of these vectors (i + 2j – 3k and 3i + $\lambda j$ + 5k). For this, we use the formula for vector product of two vectors (ai + bj + ck and pi + qj + rk) given by –
(ai + bj + ck) $\times $ (pi + qj + rk) = $\left| \begin{align}
  & i\text{ }j\text{ }k \\
 & a\text{ }b\text{ }c \\
 & p\text{ }q\text{ }r \\
\end{align} \right|$
In our case, we have,
= \[\left( i+2j3k \right)\times \left( 3i+~\lambda j+5k \right)\] = $\left| \begin{align}
  & i\text{ }j\text{ }k \\
 & \text{1 2 -3} \\
 & \text{3 }\lambda \text{ 5} \\
\end{align} \right|$ = $i(10+3\lambda )-j(5+9)+k(\lambda -6)$ -- (1)
Now, coming back to expression (A), we have –
= (2i – j – k). $\left( i(10+3\lambda )-j(5+9)+k(\lambda -6) \right)$
Now, we use the formula for scalar products, this is given by (for 2 vectors, ai + bj + ck and pi + qj + rk)
= (ai + bj + ck). (pi + qj + rk) = ap + bq + cr
In our case, we have,
= $2(10+3\lambda )+14+(6-\lambda )$
= 20 + 14 + 6 + $5\lambda $
= 40 + $5\lambda $ -- (2)
Now, since for the condition of coplanarity, the scalar triple product to be 0, expression (2) should be equated to 0. Thus, we have,
$\Rightarrow $ 40 + $5\lambda $ = 0
$\Rightarrow $ $\lambda $ = -8
Hence, the correct option is (d) -8.

Note: Another way to find the condition of coplanarity is by using the cyclic property of triple scalar. Thus, according to this property, product finding scalar triple product between the vectors $\overrightarrow{b},\overrightarrow{c},\overrightarrow{a}$ or $\overrightarrow{c},\overrightarrow{a},\overrightarrow{b}$ instead of $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$. Thus, in both these alternative ways, we will get the same answer.